Problem Description

You are given two undirected trees with n and m nodes respectively. The first tree has nodes labeled from [0, n-1] and the second tree has nodes labeled from [0, m-1].

The trees are represented by two 2D arrays:

• edges1 of length n-1 representing edges in the first tree, where edges1[i] = [ai, bi] means there's an edge between nodes ai and bi
• edges2 of length m-1 representing edges in the second tree, where edges2[i] = [ui, vi] means there's an edge between nodes ui and vi

A key concept in this problem is the "target" relationship: Node u is target to node v if the number of edges on the path from u to v is even. Note that every node is always target to itself (0 edges, which is even).

Your task is to determine, for each node i in the first tree, what is the maximum possible number of nodes that can be target to node i if you connect exactly one node from the first tree to one node from the second tree. This connection creates a combined graph where you need to count target nodes.

The output should be an array of n integers where answer[i] represents the maximum possible number of target nodes for node i in the first tree.

Important: Each query is independent - after calculating the answer for one node, the added edge is removed before processing the next node.

The solution leverages the fact that nodes at even distances from a reference point will have the same parity (both even or both odd depths), and connecting trees optimally means choosing connections that maximize the number of nodes with matching parity relationships.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The problem involves two undirected trees, which are special types of graphs. We have nodes and edges explicitly defined in edges1 and edges2.

Is it a tree?

• Yes: The problem explicitly states we're working with two undirected trees. A tree is a connected acyclic graph with n nodes and n-1 edges.

DFS

• Yes: We arrive at DFS as the recommended approach for tree problems.

Why DFS is appropriate for this problem:

1. Tree Traversal: We need to traverse both trees to understand their structure and calculate distances between nodes.

2. Distance Parity Calculation: The problem requires determining whether paths have even or odd lengths. DFS allows us to track the depth/distance of each node from a starting point, which directly relates to path parity.

3. Bipartite-like Classification: The solution uses DFS to classify nodes into two groups based on their depth parity (even or odd depth from root). This is similar to bipartite graph coloring, where DFS is the standard approach.

4. Efficient Tree Exploration: DFS efficiently explores all nodes in both trees exactly once, calculating the required parity information in O(n) and O(m) time respectively.

Conclusion: The flowchart correctly leads us to DFS for this tree-based problem. The DFS pattern is used to:

• Traverse each tree and assign parity values (0 or 1) to nodes based on their depth
• Count how many nodes have each parity value in both trees
• Use these counts to determine the maximum number of target nodes for each node in the first tree

Intuition

Let's think about what happens when we connect two trees. When we add an edge between node x from tree 1 and node y from tree 2, we create paths between all nodes in both trees.

The key insight is understanding the "target" relationship - node u is target to node v if the path between them has an even number of edges. This is essentially asking: which nodes are at even distances from each other?

In a tree, we can classify nodes into two groups based on their distance parity from any reference node. If we pick node 0 as reference and do a DFS:

• Nodes at even distances (0, 2, 4, ...) form one group
• Nodes at odd distances (1, 3, 5, ...) form another group

Here's the crucial observation: within a single tree, all nodes in the same parity group are at even distances from each other, and nodes in different groups are at odd distances from each other.

When we connect two trees by adding an edge between node i from tree 1 and node j from tree 2:

• If the edge connects nodes of the same parity (both even or both odd depth), then nodes of the same parity across both trees will be at even distances
• If the edge connects nodes of different parity, then nodes of opposite parity across trees will be at even distances

For any node i in tree 1, the target nodes come from:

1. Nodes in tree 1 that have the same parity as node i (these are always at even distance regardless of how we connect to tree 2)
2. Nodes from tree 2 that we can make "target" by choosing the right connection point

To maximize the total, we should connect to tree 2 in a way that adds the maximum possible nodes from tree 2. Since tree 2 has two parity groups, we want to connect such that the larger group becomes target to node i.

This leads to our approach:

• Use DFS to classify nodes in both trees by parity (0 or 1)
• Count how many nodes belong to each parity group in both trees
• For each node i in tree 1, the answer is: (nodes with same parity as i in tree 1) + (maximum parity group size in tree 2)

The beauty of this solution is that it reduces a complex path-counting problem to a simple parity classification problem using DFS.

Learn more about Tree, Depth-First Search and Breadth-First Search patterns.

Solution Approach

The solution implements the parity-based approach using DFS to classify nodes in both trees.

Step 1: Build Adjacency Lists

The build function converts the edge lists into adjacency list representations for both trees:

def build(edges: List[List[int]]) -> List[List[int]]:
    n = len(edges) + 1  # [Tree](/problems/tree_intro) has n-1 edges for n nodes
    g = [[] for _ in range(n)]
    for a, b in edges:
        g[a].append(b)
        g[b].append(a)
    return g

Step 2: DFS for Parity Classification

The dfs function traverses the tree and assigns parity values to each node:

def dfs(g: List[List[int]], a: int, fa: int, c: List[int], d: int, cnt: List[int]):
    c[a] = d           # Assign parity value (0 or 1) to current node
    cnt[d] += 1        # Increment count for this parity group
    for b in g[a]:
        if b != fa:    # Avoid going back to parent
            dfs(g, b, a, c, d ^ 1, cnt)  # Flip parity for child (0→1 or 1→0)

Key parameters:

• g: adjacency list of the tree
• a: current node
• fa: parent node (to avoid cycles)
• c: array storing parity value for each node
• d: current depth parity (0 or 1)
• cnt: array counting nodes in each parity group

Step 3: Main Algorithm

g1 = build(edges1)  # Build adjacency list for [tree](/problems/tree_intro) 1
g2 = build(edges2)  # Build adjacency list for tree 2
n, m = len(g1), len(g2)

c1 = [0] * n  # Parity values for [tree](/problems/tree_intro) 1 nodes
c2 = [0] * m  # Parity values for tree 2 nodes
cnt1 = [0, 0]  # Count of nodes with parity 0 and 1 in tree 1
cnt2 = [0, 0]  # Count of nodes with parity 0 and 1 in tree 2

dfs(g2, 0, -1, c2, 0, cnt2)  # Classify [tree](/problems/tree_intro) 2 nodes
dfs(g1, 0, -1, c1, 0, cnt1)  # Classify tree 1 nodes

Step 4: Calculate Results

t = max(cnt2)  # Maximum parity group size in [tree](/problems/tree_intro) 2
return [t + cnt1[c1[i]] for i in range(n)]

For each node i in tree 1:

• c1[i] gives its parity (0 or 1)
• cnt1[c1[i]] gives the count of nodes with the same parity in tree 1
• t is the maximum possible nodes we can get from tree 2
• Total target nodes = nodes from tree 1 with same parity + maximum from tree 2

Time Complexity: O(n + m) where n and m are the sizes of the two trees. We perform DFS once on each tree.

Space Complexity: O(n + m) for storing the adjacency lists and parity arrays.

The elegance of this solution lies in recognizing that the problem reduces to counting parity groups, transforming a complex graph distance problem into a simple counting problem using DFS.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Example Input:

• Tree 1: edges1 = [[0,1], [0,2], [2,3], [2,4]] (5 nodes)
• Tree 2: edges2 = [[0,1], [0,2], [0,3]] (4 nodes)

Tree 1:          Tree 2:
    0                1
   / \               |
  1   2              0
     / \            / \
    3   4          2   3

Step 1: Build Adjacency Lists

• g1: {0: [1,2], 1: [0], 2: [0,3,4], 3: [2], 4: [2]}
• g2: {0: [1,2,3], 1: [0], 2: [0], 3: [0]}

Step 2: DFS for Parity Classification

For Tree 2 (starting from node 0 with parity 0):

• Node 0: parity = 0 (depth 0)
• Node 1: parity = 1 (depth 1)
• Node 2: parity = 1 (depth 1)
• Node 3: parity = 1 (depth 1)
• Result: c2 = [0,1,1,1], cnt2 = [1,3] (1 node with parity 0, 3 nodes with parity 1)

For Tree 1 (starting from node 0 with parity 0):

• Node 0: parity = 0 (depth 0)
• Node 1: parity = 1 (depth 1)
• Node 2: parity = 1 (depth 1)
• Node 3: parity = 0 (depth 2)
• Node 4: parity = 0 (depth 2)
• Result: c1 = [0,1,1,0,0], cnt1 = [3,2] (3 nodes with parity 0, 2 nodes with parity 1)

Step 3: Calculate Maximum from Tree 2

• t = max(cnt2) = max(1,3) = 3

Step 4: Calculate Results for Each Node

For node 0 in Tree 1:

• c1[0] = 0 (parity 0)
• Nodes in Tree 1 with same parity: cnt1[0] = 3 (nodes 0,3,4)
• Maximum from Tree 2: t = 3
• answer[0] = 3 + 3 = 6

For node 1 in Tree 1:

• c1[1] = 1 (parity 1)
• Nodes in Tree 1 with same parity: cnt1[1] = 2 (nodes 1,2)
• Maximum from Tree 2: t = 3
• answer[1] = 2 + 3 = 5

For node 2 in Tree 1:

• c1[2] = 1 (parity 1)
• Nodes in Tree 1 with same parity: cnt1[1] = 2 (nodes 1,2)
• Maximum from Tree 2: t = 3
• answer[2] = 2 + 3 = 5

For node 3 in Tree 1:

• c1[3] = 0 (parity 0)
• Nodes in Tree 1 with same parity: cnt1[0] = 3 (nodes 0,3,4)
• Maximum from Tree 2: t = 3
• answer[3] = 3 + 3 = 6

For node 4 in Tree 1:

• c1[4] = 0 (parity 0)
• Nodes in Tree 1 with same parity: cnt1[0] = 3 (nodes 0,3,4)
• Maximum from Tree 2: t = 3
• answer[4] = 3 + 3 = 6

Final Answer: [6, 5, 5, 6, 6]

Why This Works:

• When we want to maximize target nodes for node i, we need nodes at even distances from i
• In Tree 1, nodes with the same parity as i are always at even distances (they're "free" targets)
• From Tree 2, we can choose to connect in a way that makes either parity group 0 or group 1 be at even distances from node i
• Since we want to maximize, we always choose the larger group from Tree 2 (which is 3 nodes with parity 1)
• The total is the sum of same-parity nodes in Tree 1 plus the maximum group from Tree 2

Solution Implementation

Time and Space Complexity

Time Complexity: O(n + m)

The algorithm consists of the following operations:

• Building graph g1 from edges1: iterates through n-1 edges, taking O(n) time
• Building graph g2 from edges2: iterates through m-1 edges, taking O(m) time
• DFS traversal on g2: visits each of the m nodes exactly once, taking O(m) time
• DFS traversal on g1: visits each of the n nodes exactly once, taking O(n) time
• Finding maximum of cnt2: O(1) since cnt2 has only 2 elements
• Building the result list: iterates through n nodes with O(1) work per node, taking O(n) time

Total time complexity: O(n) + O(m) + O(m) + O(n) + O(1) + O(n) = O(n + m)

Space Complexity: O(n + m)

The algorithm uses the following space:

• Adjacency list g1: stores n nodes with a total of 2(n-1) edges (each edge appears twice), taking O(n) space
• Adjacency list g2: stores m nodes with a total of 2(m-1) edges, taking O(m) space
• Arrays c1 and c2: O(n) and O(m) space respectively
• Arrays cnt1 and cnt2: O(1) space each (fixed size of 2)
• DFS recursion stack: at most O(n) for g1 and O(m) for g2 in the worst case (when the tree is a chain)
• Result array: O(n) space

Total space complexity: O(n) + O(m) + O(n) + O(m) + O(1) = O(n + m)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Tree Root Selection Assumption

The Pitfall: The most critical pitfall in this solution is assuming that we can arbitrarily choose node 0 as the root for both trees when performing DFS coloring. While this works for tree2 (since we're only interested in the maximum count of either color group), it can produce incorrect results for tree1 because the parity classification of nodes depends on which node is chosen as the reference point.

Why It Happens:

• When we color tree1 starting from node 0, we're essentially measuring distances from node 0
• But the problem asks for the maximum target nodes for each node i, meaning node i should be the reference point
• The parity relationship between nodes changes depending on the reference node

Example Scenario:

Tree1: 0 -- 1 -- 2 -- 3

• If we color from node 0: colors = [0, 1, 0, 1]
• If we color from node 1: colors = [1, 0, 1, 0]
• If we color from node 2: colors = [0, 1, 0, 1]

When calculating target nodes for node 1, using node 0 as reference gives wrong parity relationships.

The Solution: We need to perform DFS coloring for each query node separately:

def maxTargetNodes(self, edges1: List[List[int]], edges2: List[List[int]]) -> List[int]:
    def build_adjacency_list(edges):
        num_nodes = len(edges) + 1
        adjacency_list = [[] for _ in range(num_nodes)]
        for node_a, node_b in edges:
            adjacency_list[node_a].append(node_b)
            adjacency_list[node_b].append(node_a)
        return adjacency_list

    def dfs_color_nodes(graph, current_node, parent_node, node_colors, current_color, color_counts):
        node_colors[current_node] = current_color
        color_counts[current_color] += 1
      
        for neighbor in graph[current_node]:
            if neighbor != parent_node:
                dfs_color_nodes(graph, neighbor, current_node, node_colors, current_color ^ 1, color_counts)

    graph1 = build_adjacency_list(edges1)
    graph2 = build_adjacency_list(edges2)
  
    num_nodes_tree1 = len(graph1)
    num_nodes_tree2 = len(graph2)
  
    # For tree2, we can use any root since we only need the max count
    colors_tree2 = [0] * num_nodes_tree2
    color_counts_tree2 = [0, 0]
    dfs_color_nodes(graph2, 0, -1, colors_tree2, 0, color_counts_tree2)
    max_color_count_tree2 = max(color_counts_tree2)
  
    result = []
  
    # For each node in tree1, perform DFS with that node as root
    for query_node in range(num_nodes_tree1):
        colors_tree1 = [0] * num_nodes_tree1
        color_counts_tree1 = [0, 0]
      
        # Color tree1 with query_node as the root
        dfs_color_nodes(graph1, query_node, -1, colors_tree1, 0, color_counts_tree1)
      
        # Nodes with color 0 (same color as root) are at even distance
        # These are the target nodes from tree1
        target_nodes_tree1 = color_counts_tree1[0]
      
        # Total = target nodes from tree1 + max possible from tree2
        result.append(target_nodes_tree1 + max_color_count_tree2)
  
    return result

2. Optimization Consideration

While the corrected solution above works, it has O(n²) time complexity because we run DFS for each node. If this is too slow, a more sophisticated approach would involve:

• Pre-computing parity relationships between all pairs of nodes
• Or using the fact that we can derive parity relationships transitively

However, for moderate-sized trees (n ≤ 1000), the O(n²) solution should be acceptable.

3. Edge Case: Single Node Trees

Another pitfall is not handling the edge case where one or both trees have only a single node (empty edges list). The solution should handle this gracefully since a single node is always target to itself.